Problem 52 - College Algebra

Question

Convert each equation to standard form by ecompleting the square on \(x\) and \(y .\) Then graph the ellipse and give ehe location of its foci. $$4 x^{2}+9 y^{2}-32 x+36 y+64=0$$

Step-by-Step Solution

Answer

The standard form of the ellipse is \((x - 4)^{2}/5 + (y + 2)^{2}/(20/9) = 1\) and its graph can be drawn using the center \((4, -2)\) minor radius of \(\sqrt{5}\) and major radius of \(\sqrt{20/9}\). The foci of the ellipse are not real in the xy-plane as the calculation yields an imaginary result.

Step 1: Rearrange the equation

We can group the \(x\) and \(y\) terms and re-arrange the equation into the form \(ax^{2} + bx + c + ay^{2} + by + c = 0\). This gives: \(4x^2 - 32x + 9y^2 + 36y + 64 = 0\).

Step 2: Re-write the equation in 'completing the square' form

Rearrang the equation with constant term on the right, and complete the square for \(x\) and \(y\) terms. For \(x\), we add \((-32/2*4)^2 = 16\) inside the brackets and same on the right side of the equation. Similarly, for \(y\), we add \((36/2*9)^2 = 4\) inside the brackets and on the right side of the equation. This re-formulates the equation as, \(4(x - 4)^2 + 9(y + 2)^2 = 16 + 4\). Simplifying the equation gives, \(4(x - 4)^2 + 9(y + 2)^2 = 20\). Divide through by 20 to express in standard form of an ellipse: \((x - 4)^{2}/5 + (y + 2)^{2}/(20/9) = 1\). This is the standard form of an ellipse with center at \((4, -2)\), semi-major axis of \(\sqrt{20/9}\) and semi-minor axis of \(\sqrt{5}\).

Step 3: Draw the ellipse and calculate the foci

Now that we have the standard form of the ellipse, we can draw the graph. Mark the center at \((4,-2)\). Draw the major axis along the y-axis and minor axis along the x-axis with lengths equal to 2 times of semi-major and semi-minor lengths respectively. For the foci, recall that in an ellipse, the distance from the center to each focus is given by \(\sqrt{(semi-major axis)^2-(semi-minor axis)^2}\). By substituting axis values, we get distance from center to foci as \(\sqrt{(20/9) - 5}\) which simplifies to \(\sqrt{(20 - 45)/9} = \sqrt{-25/9}\). Since the square root of a negative number is imaginary, it implies that the foci are on the imaginary axis, or simply not real in the xy-plane.

Key Concepts

Completing the Square

To convert the given equation of the ellipse into its standard form, we need to complete the square for both variables, \(x\) and \(y\).
This process involves rearranging the terms and manipulating them to form perfect square trinomials.
For the ellipse equation \(4x^2 + 9y^2 - 32x + 36y + 64 = 0\), we start by separating the \(x\) and \(y\) terms:

• Group the \(x\) terms: \(4x^2 - 32x\)
• Group the \(y\) terms: \(9y^2 + 36y\)

To complete the square for each group, do the following:
1. **For \(x\):** Take half of \(-32\), square it, and adjust accordingly.
This leads to adding 16 to make it a complete square, \(4(x - 4)^2\).
2. **For \(y\):** Take half of \(36\), square it, and adjust accordingly,
leading to adding 4 to complete the square as \(9(y + 2)^2\). These adjustments are balanced on both sides of the initial equation, resulting in changed constant terms.

Standard Form of an Ellipse

After completing the square, the equation is reshaped into the standard form of an ellipse.
This standard form expresses an ellipse centered at a point \((h, k)\) with axes aligned to the coordinate axes.
The equation looks like:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]In our case, we rewrite:\(4(x - 4)^2 + 9(y + 2)^2 = 20\).
By dividing the entire equation by 20 to match the right-hand side to 1, the standard form becomes:\[\frac{(x - 4)^2}{5} + \frac{(y + 2)^2}{\frac{20}{9}} = 1\]Here, the center \((h, k)\) is \((4, -2)\).
The semi-major axis length is \(\sqrt{\frac{20}{9}}\) while the semi-minor axis is \(\sqrt{5}\).This form helps in easily graphing the ellipse and further analyzing its properties.

Ellipse Foci

An important feature of an ellipse is its foci, which are points located along the major axis.
The distance from the center to each focus (\(c\)) is found using:\[c = \sqrt{a^2 - b^2}\]Using our standard form, we identify:- \(a^2 = \frac{20}{9}\)- \(b^2 = 5\)Calculate \(c\) by substituting these into the formula:\[c = \sqrt{\frac{20}{9} - 5} = \sqrt{\frac{-25}{9}}\]Here, we encounter an issue: a negative number under the square root.Interestingly, this suggests imaginary foci, implying that a real graph with real dimensions considers different roots aligned with the configuration of the ellipse.
For practical graphing, focus on real values showing similar alignments without considering imaginary numbers.

Converting Equations

The challenge of converting quadratic or conic section equations lies in recognizing patterns and using algebraic techniques.
For ellipses, converting equations involves transforming a given form into one that reveals key information such as center, axes, and size.
The core process includes:

• Rearranging terms to prioritize symmetry and clarity.
• Applying 'completing the square' for both variables to achieve perfect square trinomials.
• Simplifying and dividing to shape it directly into the standard ellipse form.

Completing these steps does more than just shape the equation; it provides insightful data about the geometry involved as well.
Another common conversion type within conic sections concerns hyperbolas, circles, and parabolas, each having unique standard forms they can be converted into by employing similar methods.